In the given figure,ABCD is a cyclic quadrila | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) We know that angles in the same segment of a circle are equal.Therefore,$\angle BDC = \angle BAC$.Given $\angle BAC = 50^{\circ}$,so $\angle BDC =

Vedclass · Accepted Answer

$85$

Vedclass · Answer

(B) We know that angles in the same segment of a circle are equal.Therefore,$\angle BDC = \angle BAC$.Given $\angle BAC = 50^{\circ}$,so $\angle BDC = 50^{\circ}$.Now,$\angle ADC = \angle ADB + \angle BDC$.Substituting the values,$\angle ADC = 45^{\circ} + 50^{\circ} = 95^{\circ}$.Since $ABCD$ is a cyclic quadrilateral,the sum of opposite angles is $180^{\circ}$.Therefore,$\angle ABC + \angle ADC = 180^{\circ}$.$\angle ABC + 95^{\circ} = 180^{\circ}$.$\angle ABC = 180^{\circ} - 95^{\circ} = 85^{\circ}$.

In the given figure,$ABCD$ is a cyclic quadrilateral in which diagonals $AC$ and $BD$ intersect at $M.$ If $\angle BAC = 50^{\circ}$ and $\angle ADB = 45^{\circ}$,then find $\angle ABC$. (in $^{\circ}$)

Similar Questions

$A$ chord of a circle is equal to its radius. Find the angle subtended by this chord at a point in the major segment.

If two chords $AB$ and $CD$ of a circle intersect at right angles (see Fig.),prove that $\text{arc } CXA + \text{arc } DZB = \text{arc } AYD + \text{arc } BWC = \text{semicircle}$.

If $BM$ and $CN$ are the perpendiculars drawn on the sides $AC$ and $AB$ of the triangle $ABC$,prove that the points $B, C, M$ and $N$ are concyclic.

The region between an arc and the two radii,joining the centre to the end points of the arc is called a ........... .

In a circle with centre $P$,the length of a diameter is $20 \, cm$ and $AB$ is a chord of a circle at a distance $6 \, cm$ from the centre,then $AB = $ ........ $cm$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module